y = (e^-x) (cosx)^2 (x^2 + x + 1)
take ln of both sides
ln y = ln e^-x + ln (cosx)^2 + ln(x^2 + x + 1)
= -x + 2 ln(cosx) + ln(x^2 + x + 1)
now differentiate
y' / y = -1 + 2(-sinx/cosx) + (2x+1)/(x^2 + x + 1)
= -1 - 2tanx + (2x+1)/(x^2 + x + 1)
y' = y(-1 - 2tanx + (2x+1)/(x^2 + x + 1))
= [ (e^-x) (cosx)^2 (x^2 + x + 1) ] * [ -1 - 2tanx + (2x+1)/(x^2 + x + 1) ]
sure hope they don't expect us to simplify this
use logarithmic diff. to find the derivative of the function. Show steps please! so I can see how it is done. Thank you so much!
y=(e^(-x)cos^(2)(x))/(x^(2)+x+1)
2 answers
log y = log (e^-x) + log cos^2(x) - log(x^2+x+1)
log y = -x + 2log cos x - log(x^2+x+1)
1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)
y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1)
Now, you can massage that for a few more steps, to get something that pleases you
log y = -x + 2log cos x - log(x^2+x+1)
1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)
y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1)
Now, you can massage that for a few more steps, to get something that pleases you
2 answers